If $T$ is self-adjoint, $K$ is bound and $KT$ is symmetric, does $K$ commute with $T$?

Question

This is a question from an undergraduate-level course on operator theory on Hilbert spaces (this is not technically homework, it's half of one of 30 statements that us students may be asked to prove during our final exam which we where given ahead of schedule for the very purpose of not having to prove them on the spot; we were encouraged to use any and all possible methods to find a solution apart from asking the professor or his collaborator directly; in the very real possibility that he's reading this, we salute him as per usual).

Let $T:D(T) \subset H \rightarrow H $, where $H$ is a Hilbert space. If $T$ is self-adjoint (i.e. densely defined, hermitian on its domain and $D(T)=D(T^*)$) and $K$ is a bound operator on $H$ such that $KT$ is symmetric (i.e. densely defined and hermitian on its domain), show that $K \subset \{T\}'\colon=\{A\in \mathcal{B}(H)\mid A(D(T)) \subset D(T) \land \forall f \in D(T): ATf=TAf\} $

If $K$ is bound, then it has a unique extension from its domain to the whole Hilbert space, so I can always consider $K$ to be everywhere defined. From this, I can deduce that $ D(KT)=\{f \in D(T) \mid Tf \in D(K)=H\} = D(T) $. In general $ T^*K^*\subset(KT)^* $. Because $K$ is bound and $T$ is self-adjoint, I have that $ TK^*=T^*K^*=(KT)^* $, which implies that $K^*(D(T^*)) \subset D(T^*)$. From the symmetry of $KT$: $KTf=(KT)^*f, \forall f\in D(T)$, which implies that $TK^*=(KT)^*=KT$ on $D(T)$.

Now, if $K$ was self-adjoint, then the proof would be concluded. As @geetha290krm pointed out in the comments, if $T = 0$ then all the hypoteses of the proposition are satisfied (because $0$ is self-adjoint and $KT = 0$ is surely symmetric) but $K$ can be any operator, even a not self-adjoint one. So, now, the question is: how can the statement be proven?

P.S. I'm aware that there's another fairly similar question on here that was asked a week ago: that was done by a classmate of mine, but because he didn't get any answer we decided to rewrite it in a more complete form.

P.P.S. Any suggestion that would exceed the level of this course is still more than welcome

Answer 1

Consider $$K=\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}, \qquad T=\begin{pmatrix} -1&0\\ 0&1\end{pmatrix}.$$ Then $T$ is self-adjoint and $K$ is bounded. Furthermore, we have $$KT= \begin{pmatrix} 0&-1\\-1&0\end{pmatrix}, \qquad TK=\begin{pmatrix} 0&1\\1&0 \end{pmatrix}.$$ Thus, $KT$ is symmetric, but $KT\neq TK$, i.e. $K$ and $T$ do not commute.

Answer 2

In view of the statement "in the very real possibility that he's reading this, we salute him as per usual", I salute you back!

Indeed I stand corrected and the counterexample provided is correct! The exercise will be changed accordingly.

In order for statement to be true it turns out that $K=K^*$ is a sufficient condition. Actually one might simply state that $K$ ought to be equal to $K^*$ only on the set of $f\in\mathcal{H}$ such that $Kf\in D(T)$.